Answer:
a) ![P(X \geq 125) = 1-P(X](https://tex.z-dn.net/?f=%20P%28X%20%5Cgeq%20125%29%20%3D%201-P%28X%3C125%29%20%3D%201-P%28Z%3C%5Cfrac%7B125-130%7D%7B8%7D%29%20%3D%201-P%28Z%3C-0.625%29%20%3D%201-0.266%3D%200.734)
b) ![P(R \geq 5) = 1-P(R](https://tex.z-dn.net/?f=%20P%28R%20%5Cgeq%205%29%20%3D%201-P%28R%3C5%29%20%3D%201-P%28Z%3C%20%5Cfrac%7B5-10%7D%7B12.806%7D%29%20%3D1-P%28Z%3C-0.3904%29%20%3D%201-0.348%3D0.652)
c) ![P(\bar Y \geq 125) = P(Z> \frac{125-120}{5.774}) = 1-P(Z](https://tex.z-dn.net/?f=%20P%28%5Cbar%20Y%20%5Cgeq%20125%29%20%3D%20P%28Z%3E%20%5Cfrac%7B125-120%7D%7B5.774%7D%29%20%3D%201-P%28Z%3C0.866%29%20%3D%201-0.807%3D%200.193)
d)
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Let X the random variable that represent the scores for the East population, and for this case we know the distribution for X is given by:
Where
and
Let Y the random variable that represent the scores for the West population, and for this case we know the distribution for Y is given by:
Where
and
Part a
For this case we want this probability:
![P(X\geq 125)](https://tex.z-dn.net/?f=%20P%28X%5Cgeq%20125%29)
And we can use the z score given by:
![z = \frac{X -\mu}{\sigma}](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7BX%20-%5Cmu%7D%7B%5Csigma%7D)
And if we replace we got:
![P(X \geq 125) = 1-P(X](https://tex.z-dn.net/?f=%20P%28X%20%5Cgeq%20125%29%20%3D%201-P%28X%3C125%29%20%3D%201-P%28Z%3C%5Cfrac%7B125-130%7D%7B8%7D%29%20%3D%201-P%28Z%3C-0.625%29%20%3D%201-0.266%3D%200.734)
Part b
For this case we need to define the following random variable R = X-Y and we know that the distribution of R is given by:
![R \sim N (130-120= 10, \sigma_R = \sqrt{8^2 +10^2}=12.806)](https://tex.z-dn.net/?f=%20R%20%5Csim%20N%20%28130-120%3D%2010%2C%20%5Csigma_R%20%3D%20%5Csqrt%7B8%5E2%20%2B10%5E2%7D%3D12.806%29)
And we want this probability:
![P(R\geq 5)](https://tex.z-dn.net/?f=%20P%28R%5Cgeq%205%29%20)
We can use the z score given by:
![z= \frac{R -\mu_R}{\sigma_R}](https://tex.z-dn.net/?f=%20z%3D%20%5Cfrac%7BR%20-%5Cmu_R%7D%7B%5Csigma_R%7D)
If we use this formula we got:
![P(R \geq 5) = 1-P(R](https://tex.z-dn.net/?f=%20P%28R%20%5Cgeq%205%29%20%3D%201-P%28R%3C5%29%20%3D%201-P%28Z%3C%20%5Cfrac%7B5-10%7D%7B12.806%7D%29%20%3D1-P%28Z%3C-0.3904%29%20%3D%201-0.348%3D0.652)
Part c
For this case we select a sample size of n =3 for the Y distribution, the sample mean have the following distribution:
![\bar Y \sim N(120, \frac{10}{\sqrt{3}}=5.774)](https://tex.z-dn.net/?f=%20%5Cbar%20Y%20%5Csim%20N%28120%2C%20%5Cfrac%7B10%7D%7B%5Csqrt%7B3%7D%7D%3D5.774%29)
And we want this probability:
![P(\bar Y \geq 125) = P(Z> \frac{125-120}{5.774}) = 1-P(Z](https://tex.z-dn.net/?f=%20P%28%5Cbar%20Y%20%5Cgeq%20125%29%20%3D%20P%28Z%3E%20%5Cfrac%7B125-120%7D%7B5.774%7D%29%20%3D%201-P%28Z%3C0.866%29%20%3D%201-0.807%3D%200.193)
Part d
For this case we define the following random variable
and the distribution for H is given by:
![H \sim N (130-120=10, \sigma_H = \sqrt{\frac{8^2 +10^2}{3}}= 7.394)](https://tex.z-dn.net/?f=%20H%20%5Csim%20N%20%28130-120%3D10%2C%20%5Csigma_H%20%3D%20%5Csqrt%7B%5Cfrac%7B8%5E2%20%2B10%5E2%7D%7B3%7D%7D%3D%207.394%29)
And the z score would be given by:
![z = \frac{H -\mu_H}{\sigma_H}](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7BH%20-%5Cmu_H%7D%7B%5Csigma_H%7D)
And if we find the probability required we got: